Problem #5: Which of the following statements are always true for vectors in R3,

Question

Problem #5: Which of the following statements are always true for vectors in R3, (i) If u . (v × w) 4 then w . (v × u)--4 (ii) (7u + v) x (u + 6v) = 43 (u × v) iii If u is orthogonal to v and w then u is also orthogonal to liwll v llvll w (C) (i) only (D) (iii) only (E) and (ii) only (F) none of them (A) all of them (B) (ii) and (iii) only (G) () and (ii) only (H(ii only Select Problem #5: Just Save Submit Problem #5 for Grading Attempt #2 Attempt #3 Problem #51 Attempt #1 Your Answer: Your Mark:

Explanation / Answer

(i) u.(v x w) is the scalar triple product between u,v and w.

According to the properties of scalar triple product of vectors:

u.(v x w) = v.(w x u) = w.(u x v)

From the above property, w.(u x v) = 4

Now, if we change the position of two vectors in a cross product, the sign of the product changes.

For example, (a x b) = - (b x a)

Hence, w.(u x v)= w. (-(v x u))= -w.(v x u)

but w.(u x v) = 4, hence from the above relation, -w.(v x u) = 4

Therefore, w.(v x u) = -4

(ii) Just expand this like ordinary multiplication,

(7u +v) x (u+6v)= 7(u x u) +(7u x 6v) + (v x u) + (v x 6v)

=7(u x u) + 42(u x v) + (v x u) + 6(v x v) [ We can bring all the scalars outside]

= 0 + 42(u x v) + (v x u) + 0 [Cross product of a vector with itself is always 0]

= 42(u x v) - (u x v) [interchanging position of 2 vectors will lead to change in sign]

= 41(u x v)

(iii) When 2 vectors are orthogonal, their dot product is 0.

Hence, u.v = 0 and u.w = 0

||w|| is length of vector w and hence is a scalar constant

||v|| is length of vector v and hence is a scalar constant

If u is orthogonal to ||w||v + ||v||w

Then u.( ||w||v + ||v||w ) = (u.||w||v) + (u.||v||w)

= ||w||(u.v) + ||v||(u.w) [Bring all the scalar constants outside]

But u.v= u.w= 0

= ||w||(0) + ||v||(0)

= 0 + 0 = 0

Since the dot product is 0, u is orthogonal to ||w||v + ||v||w

Hence option G, that is only (i) and (iii), are correct

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